If the point P(acosθ,bsinθ) satisfies the equation a2x2+b2y2=1 , then it lies on the given ellipse .
Now a2x2+b2y2=a2(acosθ)2+b2(bsinθ)2
=a2a2cos2θ+b2b2sin2θ
=cos2θ+sin2θ
=1.
∴ point P(acosθ,bsinθ) lies on the given ellipse. (PROVED).
Now the equation of the tangent at the point P(acosθ,bsinθ) is given by
a2x∗acosθ+b2y∗bsinθ=1or, axcosθ+bysinθ=1or, abbxcosθ+aysinθ=1or, bxcosθ+aysinθ=abor, aysinθ=−bxcosθ+abor, y=−asinθbxcosθ+asinθabor, y=−(abtanθ)x+bcosecθ.
∴ Gradient of the tangent at the point P(acosθ,bsinθ) is −(abtanθ) . Answer.
Equation of Normal: We know that normal is perpendicular to the tangent..
so if equation of tangent is bxcosθ+aysinθ=ab then equation of normal is of the form
axsinθ−bycosθ=k, where k is constant.
Now as this normal is passing through P(acosθ,bsinθ) ...by putting this coordinate into the equation of normal we get the value of k .
∴a∗acosθ∗sinθ−b∗bsinθ∗cosθ=kor, k=a2sinθcosθ−b2sinθcosθor, k=(a2−b2)sinθcosθ
∴ The equation of normal at the point P(acosθ,bsinθ) is given by
axsinθ−bycosθ=(a2−b2)sinθcosθ. PROVED.
Now if the equation of normal is passing through B(0,b) then we have
a∗0∗sinθ−b∗b∗cosθ=(a2−b2)sinθcosθor, −b2cosθ=(a2−b2)sinθcosθ.or, −b2=(a2−b2)sinθor, sinθ=a2−b2−b2
Now as we know −1≤sinθ,
∴−1≤a2−b2−b2or,1≥a2−b2b2or,a2−b2≥b2or,a2≥2b2 PROVED.
Now tangent at P(acosθ,bsinθ) which is bxcosθ+aysinθ=ab , meets the Y axis at Q.
putting x=0 ,we get
0+aysinθ=abor, ysinθ=bor, y=sinθb.
∴ The coordinate of Q is (0,sinθb) .
BQ=(sinθb−b)BQ=(a2−b2−b2b−b)
BQ=b(b2b2−a2−1)BQ=b∗(b2b2−a2−b2)BQ=b−a2
∴∣BQ∣=ba2 PROVED.
After solving question i found that BQ is ba2 but in a given question it was given as b2a2 .. This is a mistake.
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